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Question

If the distance between the points (5,2) and (1,a) is 5 units, then the sum of all possible values(s) of a is

A
3
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B
1
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C
4
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D
4
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Solution

The correct option is C 4
Distance between two points
5=(51)2+(2a)2
5=16+4+a2+4a25=20+a2+4aa2+4a5=0(a+5)(a1)=0a=5,1

Hence, the sum of possible values of a is 4.

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